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SSSSS [86.1K]
3 years ago
13

What is solution set for the quadratic equation ×^2 + 10×+25=0​

Mathematics
1 answer:
Roman55 [17]3 years ago
7 0

Answer:

x=-5

Step-by-step explanation:

x squared+10x+25 factors into (x+5)(x+5). Set x equal to 0 and you get x=-5. To check, plug in negative 5 and you get 25+ -50 + 25=0 which is true.

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Can someone help me with this math homework please!
kobusy [5.1K]

Answer:

It's 2, 1, and y = 2x + 1.

Step-by-step explanation:

You can see the rise is 2 and the run is 1, making the slope = 2, and the y-intercept is 1 because that is where it crosses the y axis. Once you have the slope and y intercept, you can put it in a function, with the form being y=2x+1, the slope being the number before the x and the y-int value being after the x.

8 0
2 years ago
Read 2 more answers
When mindy went to China, she exchanged $1.00 for 6.589 yuan . What digit is in the tenths place and the hundredth place of 6.58
jenyasd209 [6]

Answer:

tenth place= 5

hundredth= 8

Step-by-step explanation:

tenth is 5 since 5÷10=0.5

6 0
3 years ago
Find the real numbers x and y if -3+ix^2y and x^2+y+4i are conjugate of each other. Pls solve with the steps
Firdavs [7]
ANSWER
x = ±1 and y = -4.
Either x = +1 or x = -1 will work

EXPLANATION
If -3 + ix²y and x² + y + 4i are complex conjugates, then one of them can be written in the form a + bi and the other in the form a - bi. In other words, between conjugates, the imaginary parts are same in absolute value but different in sign (b and -b). The real parts are the same

For -3 + ix²y
⇒ real part: -3
⇒ imaginary part: x²y

For x² + y + 4i
⇒ real part: x² + y (since x, y are real numbers)
⇒ imaginary part: 4

Therefore, for the two expressions to be conjugates, we must satisfy the two conditions. 

Condition 1: Imaginary parts are same in absolute value but different in sign. We can set the imaginary part of -3 + ix²y to be the negative imaginary part of x² + y + 4i so that the 

   x²y = -4 ... (I)

Condition 2: Real parts are the same

   x² + y = -3 ... (II)

We have a system of equations since both conditions must be satisfied

   x²y = -4 ... (I)
   x² + y = -3 ... (II)

We can rearrange equation (II) so that we have

   y = -3 - x² ... (II)

Substituting into equation (I)

   x²y = -4 ... (I)
   x²(-3 - x²) = -4
   -3x² - x⁴ = -4
   x⁴ + 3x² - 4 = 0
   (x² + 4)(x² - 1) = 0
   (x² + 4)(x-1)(x+1) = 0

Therefore, x = ±1.
Leave alone (x² + 4) as it gives no real solutions.

Solve for y:

   y = -3 - x² ... (II)
   y = -3 - (±1)²
   y = -3 - 1
   y = -4

So x = ±1 and y = -4. We can confirm this results in conjugates by substituting into the expressions:

   -3 + ix²y 
   = -3 + i(±1)²(-4)
   = -3 - 4i

   x² + y + 4i
   = (±1)² - 4 + 4i
   = 1 - 4 + 4i
   = -3 + 4i

They result in conjugates
4 0
3 years ago
Read 2 more answers
17 more than some number is 57
anastassius [24]
17+x=57.
So, subtract 17 from both sides
X=40
6 0
3 years ago
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Ten feet below sea level is farther below sea level than 2 and a half feet below sea level
bulgar [2K]

Answer:

correct

Step-by-step explanation:

5 0
3 years ago
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